How can I demonstrate that $x-x^9$ is divisible by $30$ whenever $x$ is an integer?
I know that $$x-x^9=x(1-x^8)=x(1-x^4)(1+x^4)=x(1-x^2)(1+x^2)(1+x^4)$$
but I don't know how to demonstrate that this number is divisible by $30$.
Asked
Active
Viewed 131 times
4
-
3do primes $2,3,5$ one at a time – Will Jagy Feb 25 '15 at 20:31
-
1Hint: Factor it. – Lucian Feb 25 '15 at 20:32
-
I don't know how to factor it – user112127 Feb 25 '15 at 20:37
-
@user112127 how do you not know? You wrote it factored when you said $x-x^9 = x(1-x^2)(1+x^2)(1+x^4)$ – graydad Feb 25 '15 at 20:38
-
Ok. So x−x^9=x(1-x)(1+x)(1+x^2)(1+x^4) – user112127 Feb 25 '15 at 20:39
-
1and (1-x)x(1+x) is divisible by 3 . Am i right ? – user112127 Feb 25 '15 at 20:40
-
@user112127 that depends. Are you assuming $x$ is an integer? Or can it be any number? – graydad Feb 25 '15 at 20:41
-
@ graydad x is integer. – user112127 Feb 25 '15 at 20:43
-
and x(x+1) is dibisible by 2 . Now we have to prove that the number is divisible by 5 – user112127 Feb 25 '15 at 20:48
-
@user112127 The product of three successive integers is divisible by $3!=6$: proof $\binom n3 = n(n-1)(n-2)/6$ is an integer. – Mark Bennet Feb 25 '15 at 20:49
3 Answers
7
Let's factor $x^9-x$ like you have done: $$ x^9-x=(x-1)x(x+1)(x^2+1)(x^4+1).\tag{$*$} $$ Let's look at the RHS. The product of the first 2 terms is divisible by $2$ because it consists of 2 consecutive integers. Similarly, the product of the first 3 terms is divisible by $3$. Now, if you had $$ (x-2)(x-1)x(x+1)(x+2) $$ then of course that would be divisible by $5$ as well. But note this $$ (x-1)x(x+1)(x^2+1)-(x-2)(x-1)x(x+1)(x+2)=5x(x^2-1)\equiv 0\pmod{5}. $$ So the product of the first 4 terms of the RHS of ($*$) is also divisible by $5$. Now you're done.
Kim Jong Un
- 15,004
- 1
- 26
- 50
-
-
You're welcome. The divisibility by $5$ trick above comes from noticing $x^2+1=5+(x^2-4)=5+(x-2)(x+2)$. – Kim Jong Un Feb 25 '15 at 20:53
-
-
6
You have to prove that $x-x^9$ is divisible by $2$, $3$ and $5$.
$x\equiv x^9\pmod{2}$ is obvious, isn't it?
By Fermat's little theorem, $x^3\equiv x\pmod{3}$, so $x^9=(x^3)^3\equiv x^3\equiv x\pmod{3}$
By Fermat's little theorem, $x^5\equiv x\pmod{5}$, so $x^9=x^4x^5\equiv x^4 x\equiv x^5\equiv x\pmod{5}$.
egreg
- 244,946
0
It's a special case of the following theorem. The direction we need $(\Leftarrow)$ has a very simple proof.
Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$
$\qquad\rm n\:|\:a^e-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\:\Rightarrow\: p\!-\!1\:|\:e\!-\!1$
Bill Dubuque
- 282,220